# Ring the Bell again

1. Jun 21, 2014

### lrhorer

There are several threads on the Bell paradox, plus the article in the FAQ forum, but I must be missing something here.

Forget for a moment about 2 ships. Let's take one ship, which an observer at the front and the other at the rear. The ship is undergoing a constant 1G acceleration. The conclusion from the proposed paradox would seem to be the compressive stress in the skin of the ship is less than 1G, and that at some point in time it may actually fall to zero and even eventually become an expansive stress and perhaps even grow so large that the ship is ripped apart. I don't see how that would be possible, since at any point in time both observes and the ship itself are all at rest WRT each other, irrespective of the ships motion wrt any external frame of reference. As I say, I'm missing something, here. Someone please explain.

We have two identical ships, pined in the same direction, a few hundred meters or so apart. Ship 1 begins accelerating toward ship 2 at time t1, so that an observer in the ship feels a constant 1G acceleration. Ship 2 begins accelerating at time t2, away from ship 1. Clearly there are an infinite number of values of t1 and t2 that will allow ship 1 to overtake ship 2. There are likewise an infinite number of values of t1 and t2 that will cause ship 2 to out-distance ship 1. The proposed paradox is suggesting, however, there are no values for t1 and t2 that would ever allow the two ships to remain motionless wrt each other. Again, someone please explain this to me.

OK, now to the really fun stuff. In stead of rockets, lets talk about a circular disk. The disk has a small mass, but infinitely strong, so it won't fly apart. It is not, however, rigid. It can be deformed under stress. There is a straight painted along a diameter of the disk. We start spinning the disk with a constant angular acceleration measured at the edge of the disk. The straight line takes on a very slight curve shaped rather like an S. At normal velocities the shape of the curve does change radically as the disk's rotational velocity increases, because that the disk grows larger in diameter with each passing second, and because the torque required to maintain the angular acceleration increases. Nonetheless, the angular velocity of the edge of the disk is the same as that toward the center at any given time, and the curve is only changing slightly with time. The question: What happens as the outer edge of the disk approaches the speed of light? What happens to the curved line?

2. Jun 21, 2014

### WannabeNewton

This is not true. If you are taking the ship to be a one-dimensional extended object and give every point of the space a proper acceleration of g then it is very easy to show that in the local frame of any one point of the ship the neighboring points of the ship have non-vanishing radial velocities relative to this frame. You are assuming that the ship remains rigid during the acceleration which is an incorrect assumption. There is no such thing as a rigid object in relativity. We only have rigid motion which is exactly defined by the vanishing of radial velocities of neighboring points relative to the local frame of some given point. For one-dimensional objects or a family of one-dimensional objects, a constant proper acceleration that is uniform throughout the object or family of objects will fail to produce rigid motion. The reason for this is given quite clearly and explicitly in the Bell Paradox FAQ.

See above.

No it doesn't. If the torque is applied tangent to the rim of the disk then the ensuing deformation is parallel to the circumference of the disk and stretches the circumference, that's all. The mark on the rim of the disk will get displaced as a result.

What?

Again...what? If you mean that the angular velocity of the disk, as measured relative to an inertial observer at the center, is the same at every point then this is only true for rigid disks.

The mark on the rim of the disk will keep getting displaced from the rest of the line as the circumference gets stretched, until the stresses breakup the disk.

3. Jun 21, 2014

### lrhorer

No, I am not. I am merely asserting that the force exerted on the frame of the ship is constant. It never changes. Suppose the ship has been accelerating for, oh, 1000 years or so. How and why do the forces on the skin of the ship ever change significantly? How is the forward observer (or any point on the ship) traveling at a different velocity than any other point on the ship? Through what auspices? After all, their relative velocity has never changed above zero. How are the local spacetime properties of the ship any different than one which has only been accelerating for a few seconds?

I believe I understand that, at least somewhat.

Not to me, it isn't. At what rate do their velocities vary for a given acceleration? How and why is the force accelerating the forward observer greater than that accelerating the rear observer? What is the source of the additional kinetic energy imparted to the forward observer?

You also never answered my question concerning values of t1 and t2 for the independently accelerating ships. We most surely can produce values for t1 and t2 that produce a vanishingly small closure rate for the two ships as well as values that can produce vanishingly small increases of the distance between the two. The inference from the paradox is there is no value of t1 and t2 that can provide that the two remain at a fixed distance, despite our being able to arrange for arbitrarily low rates of closure or separation. Again, please explain.

You didn't read what I wrote. The mark is not "on" the rim. It spans the disk, going from the rim on one side, through the center, and winds up on the other side. An ordinary torque applied to the shaft passing through the center will cause the straight line segment to bend into a shallow S curve even when the disk is only rotating at a few RPM.

Surely you must know an increase in angular velocity of a flexible disk will cause the diameter of the disk to increase? Have you never heard of a centrifugal clutch, used commonly in chain saws, weed eaters, wood chippers, and other gasoline powered tools? Have you never seen the wheels of a drag racer coming off the starting line?

Which this is not. That's the whole point.

The line is not on the rim, so this conclusion is simply incorrect. The line is across the disk. At sub-relativistic speeds, the line only changes slowly as the geometry of the disk changes solely due to stresses in the disk. The S curve merely gets very slowly more curvaceous and longer. How do we describe its change in shape with time as the edge crosses over into relativistic velocities? A madly tightening spiral? The winding of an insane watch spring? I don't think that is quite correct.

Last edited: Jun 21, 2014
4. Jun 21, 2014

### Staff: Mentor

The scenario you have outlined here is almost entirely unrelated to the scenario in the Bell paradox. I am not certain why you think that it would make any conclusion about a substantively different scenario.

In Bell's the distance is constant in the initial frame. In yours the distance is constant in the momentarily co moving inertial frame. Different scenarios lead to different conclusions.

5. Jun 21, 2014

### Staff: Mentor

Yes, and you are failing to understand the implications of this specification in relativity. They are *not* the same as they would be in Newtonian physics; that's the point of the Bell spaceship paradox.

For the parts of the ship to all remain at rest relative to each other, the force felt at the front of the ship *cannot* be the same as the force felt at the rear; the latter *must* be larger. This is a straightforward implication of relativistic kinematics. An immediate corollary of this is that, if the force felt at the front of the ship *is* exactly the same as the force felt at the rear, the ship will stretch and eventually be torn apart; this is similar to the conclusion of the Bell spaceship scenario.

They don't. What changes is not the force on each end of the ship--by hypothesis, those are constant and exactly the same everywhere on the ship. What changes is the relative velocity between the front and rear of the ship.

Because that's how relativistic kinematics works. You are implicitly using Newtonian, non-relativistic kinematics in your reasoning, and that is leading you astray. If the explanation in the FAQ is not helping you, I would recommend working through the underlying math in detail yourself; that will also help to familiarize you with the differences between the correct relativistic kinematics and the incorrect things your Newtonian intuition is telling you.

6. Jun 21, 2014

### Staff: Mentor

I think that when he says "the ship is undergoing a 1G acceleration" he means literally that every single piece of the ship has a proper acceleration of exactly 1G. That means the congruence of worldlines describing the ship is a Bell congruence, not a Rindler congruence. (More precisely, I don't think he understands that he can't specify both that every part of the ship accelerates at exactly 1G, *and* that all the parts of the ship remain at rest relative to each other; in relativity there is *no* congruence of worldlines that satisfies both properties.)

7. Jun 21, 2014

### Staff: Mentor

Hmm, it is possible that is what he meant. I interpreted it as a more ordinary specification of a traditional steel ship being pushed by a traditional 1 g thrust from the rear.

8. Jun 22, 2014

### stevendaryl

Staff Emeritus
In a rocket moving at relativistic speeds, if the rear of the rocket is undergoing 1G acceleration, then the front of the rocket will be accelerating at a slightly smaller rate. That is, assuming that it's a normal rocket, where the force is applied to the rear, and that force is conveyed to the rest of the ship through contact.

Of course, you can imagine putting boosters on both the front and rear of the rocket, to make sure that they are both accelerated at the same rate. But if you do that, you're essentially back in the case of two rockets (the original Bell paradox). If the boosters are powerful enough, they will rip the rocket in two pieces.

9. Jun 22, 2014

### Staff: Mentor

Something that underlies much of this discussion but hasn't been stated explicitly but might help OP: whenever we speak of a uniformly accelerating object (as opposed to an idealized point) there's a subtle relativity of simultaneity issue.

When you say that the nose and the tail of the ship are both accelerating "at the same rate" or "uniformly", you're basically saying that they're changing their speed by the same amount at the same time... And any time you see that phrase, you know you need to be more careful about what you're specifying.

10. Jun 22, 2014

### lrhorer

Well, not entirely, but I see the point that was being made, now. Really, it boils down to answering the question I asked in the first place. The question was, "Are there values for t1 and t2 that allow the spaceships to remain at a fixed distance." The answer is, "No", and it makes sense once that question is answered.

11. Jun 22, 2014

### lrhorer

That has been discussed in the other threads, but to my mind stating it that way obscures what is going on. A simple simultaneity issue (yes, I know there isn't really such a thing) makes it sound as if in fact there are two values of t1 and t2 that will allow the two ships to remain at a fixed distance. There aren't. That fact makes the situation clear to me.

Yes, but I don't believe I ever said that. To my way of understanding - and I did not see this mentioned explicitly in the other conversations - the crux of the matter is that a pair of objects separated in space along the direction of motion under the exact same acceleration do not see the space between themselves the same way. Looking back I see it was indeed discussed, but not in a way that was clear to me. Obviously it is clearer now, and frankly I should have seen it before. I guess I am getting old. Each of the ships sees the universe compressed looking forward, and stretched looking backwards. Two ships side by side don't have the same issue.

Last edited: Jun 22, 2014
12. Jun 22, 2014

### lrhorer

Your interpretation of my scenario is correct. Of course, the other scenario is also interesting. I am pedestrian enough to enjoy seeing things ripped apart.

13. Jun 22, 2014

### stevendaryl

Staff Emeritus
Well, if you are accelerating the front and the rear of the rocket separately (which you would have to do, if you wanted them both to experience the same acceleration), then you're treating them as two different objects, so the scenario is not much different from the Bell Paradox case of two rockets connected by a string. Instead, the two objects are connected by steel walls, but it's the same principle.

14. Jun 22, 2014

### lrhorer

To my mind, that is a trivial aspect of the illustration. That almost nothing is the same in Newtonian physics as in Relativity is patently obvious. To my way of thinking, the important thing to understanding the issue is the fact there is a difference to the universe along the direction of acceleration than perpendicular to it. Now, I already knew that, actually, but all the other discussions were obscuring the connection in my understanding.

Yes, I know, "I told you so" applies, but in my defense, presuming I need one, it has been over 30 years since I looked at any examples in Relativity in any more than cursory way.

Extremely similar, yes. To be sure, it is precisely the same principle.

Yes, because the distance changes. Space along that axis is no longer the same as it was prior to the acceleration being applied.

Never bad advice, but the fact is a simple answer of, "No", to my question would have sufficed.

15. Jun 22, 2014

### lrhorer

Precisely.

16. Jun 22, 2014

### lrhorer

I don't see why you seem to be saying it is unimportant. Precisely the same fundamental principles apply to both situations. Understanding how and why those principles result in vastly different results under nearly - but not quite - identical circumstances is crucial to a proper understanding of how the principles work in general. The best way to come to understand one's own language is to learn a different one. One of the best ways to understand fully how a rifle works is to study a pistol. One of the best ways, if you ask me, to understand what is happening on two separate ships is to look at how the situation is different on a single ship of non-zero length.

Sometimes they do and sometimes they don't. When they do, then one would infer the general principles are acting the same way under two different scenarios. This helps gain understanding as well. Example: In the first scenario we have a pair of trucks traveling on a road toward each other. Both are traveling at 50 MPH. In the second scenario, only one truck is moving, but at 100 MPH. In the third scenario, we only have one truck, traveling at 50MPH, but it is hitting a hypothetically immovable wall. Ignoring any effects from the pavement, how are the forces applied to any one of the trucks different in the three scenarios? Answer: They aren't. This is very illustrative.

On the other hand, if we take a massive circular disk at rest with an axle passing through its center and we apply a torque perpendicular to the axle we get one result, while if the disk is rotating we get a very different result. This is also very illustrative, but of a different principle.

Similarly, if the two rockets are perfectly side-by side, the result is precisely as one would expect from an SR perspective, or even a Newtonian perspective, for that matter. If the two are displaced along the axis of acceleration, it is a different matter. Whether the two observers are being accelerated independently or are linked in a special manner does not change the underlying principles, but it does change the calculations. That is the crux of the matter, and understanding the fundamental differences between the three scenarios is crucial. For me, at least, it is also frequently the fastest and easiest way to learn.

Last edited: Jun 22, 2014
17. Jun 22, 2014

### lrhorer

Who says it is moving at Relativistic speeds? And relative to what? This scenario does not require the ships to be moving at anything more than a vanishingly small velocity WRT the initial external reference frame.

Yes, but the effect is instantaneous. The elapsed time from when the force is initially applied can approach zero, and the ships' velocities WRT an un-accelerated observer can be arbitrarily small.

18. Jun 22, 2014

### lrhorer

Quite. I would definitely enjoy seeing a rocket being ripped apart relativistically - as long as no one was aboard.

It is relevant to note, I think, that reproducing the experiment at 1G would be impossible in practical terms. It just would not be possible to attach a rope frail enough to break under the tiny differential force which would exist at 1G without it breaking from handling. If we had, say, a rail gun that applied its acceleration perfectly evenly across the entire projectile and was capable of producing 1 Billion G or some such, we might see the projectile ripped apart by the Relativistic forces.

Last edited: Jun 22, 2014
19. Jun 22, 2014

### Staff: Mentor

I didn't say anything about the importance of the question. My point was not to denigrate the importance of your question, but simply to point out the fact that you are taking the outcome of Bell's paradox and attempting to apply the conclusions to a substantively different scenario. That is a fundamentally flawed approach.

Correct, you need to apply the same principles to analyze both situations, not merely try to paste the conclusions of one scenario onto the facts of the other.

In this case the different conclusions for the different scenarios can all be obtained through the principles of the Lorentz transform and the proper acceleration. If you work through the math then you can take any pair of worldlines, determine the proper acceleration on each, and Lorentz transform to determine the distance between them in any inertial frame including a momentarily co-moving inertial frame.

20. Jun 22, 2014

### WannabeNewton

Simultaneous equal acceleration in the lab frame = non-simultaneous acceleration in the local frame of any given point of the ship. That's all there is to it.

It has already been explained multiple times. Seriously what more could possibly be said? If you want the ships to remain at a constant distance relative to one another in each of their rest frames then the front ship must accelerate with a greater magnitude than the rear ship. If you want them to remain the same distance in the lab frame then the same acceleration suffices. Your question isn't even worded properly because you've failed to indicate in what frame the ships close in on one another or remain at a constant distance.

What in the world does that have to do with the Bell paradox? Those are just garden variety shearing and centrifugal effects respectively. If you want to ask a totally separate question then go ahead but don't imbue it in a completely different one. If you only focus on relativistic effects, as I assumed you were, then the only effect of the torque applied tangential to the rim is to stretch the circumference (more precisely, stretch infinitesimal rulers laid out along the circumference) due to the purely relativistic notion of Born rigidity (or lack thereof in this case). Try to make your posts less incoherent next time so that people can actually read into whatever it is you're trying to convey. Based on your other threads you seem to have a problem with this.

21. Jun 22, 2014

### Staff: Mentor

Yes, I did say "almost entirely". In Bell's scenario the front and back have the same coordinate acceleration, in yours they do not. In Bell's scenario the front and back do not move rigidly, in yours they do (neglecting initial transients). In Bell's scenario the proper acceleration is the same in front and back, in yours it is not.

But both scenarios are involving spaceships accelerating relativistically in the absence of gravity.

More important than answering the question is understanding how to determine the answer for the question. Are you comfortable with that? If you would like to work it out then I would be glad to help.

22. Jun 22, 2014

### stevendaryl

Staff Emeritus
I don't quite understand what you are driving at. It might be instructive (I really hesitate about doing this, since it might cause more confusion than it's worth) to consider what things look like from the point of view of someone aboard an accelerating rocket.

It's possible for the observer to pick a coordinate system (the Rindler coordinates) such that life on board the rocket seem very similar (qualitatively, but not quantitatively) to life on a rocket that is hovering above the surface of a planet. Some examples of phenomena:

1. If you are standing on the floor of the rocket and drop an object, it will fall to the floor.
2. The precise rate at which the dropped object appears to accelerate downward depends on the height from which it is dropped--the "acceleration due to gravity" is smaller at greater heights.
3. There are other effects that don't come into play in the Bell spaceship scenario, such as "gravitational time dilation": clocks closer to the floor run slightly slower than clocks higher up.

As I said, in detail, the two situations are slightly different--the variation of the apparent gravitational acceleration with height is different in the two cases. But qualitatively, the situations are very similar.

Now, as I said, a rocket, whether it is hovering in place over a planet or accelerating through space, will experience slightly different accelerations at the bottom and top of the rocket. The only way to get the top of the rocket to experience the same acceleration as the bottom of the rocket is to literally break the rocket in half, and accelerate the top part separately from the bottom part. In particular, the top has to accelerate away from the bottom in order for both to "feel" the same acceleration.

23. Jun 22, 2014

### Staff: Mentor

That's actually not the question that the Bell spaceship paradox gives a "No" answer to. The Bell spaceship paradox just says that the answer to "Will the ships remain at rest relative to each other if t1 = t2?" is "No". It says nothing about what happens if t1 does not equal t2. The latter case requires a more general analysis, which I'll try to post when I have more time.

24. Jun 22, 2014

### stevendaryl

Staff Emeritus
So your original question was about the timing of launching of two different ships, one ahead of the other. It doesn't matter what the timing is, if the rate of acceleration "felt" by the front ship is equal to the rate felt by the rear ship, then the front ship will pull away from the rear ship (as measured by those on board the two ships).

25. Jun 22, 2014

### Staff: Mentor

I don't think this is always true; for example, if the front ship launches long enough after the rear ship, the rear ship can pass the front ship, and once that happens, the "rear" ship is now in front and it is the one that will pull away. As I said in my response to the OP just now, a more general analysis is needed for the case where the ship launch times are allowed to be different (with respect to the original rest frame of both of them).